Difference between revisions of "1995 IMO Problems/Problem 1"
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== Hint == | == Hint == | ||
− | Radical | + | Think about Radical Axis, Power of a Point and Radical Center. |
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− | |||
== Solution == | == Solution == | ||
− | Since <math>M</math> is on the circle with diameter <math>AC</math>, we have <math>\angle AMC=90</math> and so <math>\angle MCA=90-A</math>. We | + | Since <math>M</math> is on the circle with diameter <math>AC</math>, we have <math>\angle AMC=90</math> and so <math>\angle MCA=90-A</math>. We similarly find that <math>\angle BND=90</math>. Also, notice that the line <math>XY</math> is the radical axis of the two circles with diameters <math>AC</math> and <math>BD</math>. Thus, since <math>P</math> is on <math>XY</math>, we have <math>PN\cdot PB=PM\cdot PC</math> and so by the converse of Power of a Point, the quadrilateral <math>MNBC</math> is cyclic. Thus, <math>90-A=\angle MCA=\angle BNM</math>. Thus, <math>\angle MND=180-A</math> and so quadrilateral <math>AMND</math> is cyclic. Let the name of the circle <math>AMND</math> be <math>O</math> . Then, the radical axis of <math>O</math> and the circle with diameter <math>AC</math> is line <math>AM</math>. Also, the radical axis of <math>O</math> and the circle with diameter <math>BD</math> is line <math>DN</math>. Since the pairwise radical axes of 3 circles are concurrent, we have <math>AM,DN,XY</math> are concurrent as desired. |
==Solution 2== | ==Solution 2== | ||
− | Let <math>AM</math> and <math>PT</math> intersect at <math>Z</math>. Now, assume that <math>Z, N, P</math> are not collinear. In that case, let <math>ZD</math> intersect the circle with diameter <math>BD</math> at <math>N'</math> and the circle through <math>D, P, T</math> at <math>N''</math>. | + | Let <math>AM</math> and <math>PT</math> (a subsegment of <math>XY</math>) intersect at <math>Z</math>. Now, assume that <math>Z, N, P</math> are not collinear. In that case, let <math>ZD</math> intersect the circle with diameter <math>BD</math> at <math>N'</math> and the circle through <math>D, P, T</math> at <math>N''</math>. |
− | We know that <math> | + | We know that <math>\angle AMC = \angle BND = \angle ATP = 90^\circ</math> via standard formulae, so quadrilaterals <math>AMPT</math> and <math>DNPT</math> are cyclic. Thus, <math>N'</math> and <math>N''</math> are distinct, as none of them is <math>N</math>. Hence, by Power of a Point, <cmath>ZM * ZA = ZP * ZT = ZN'' * ZD.</cmath> However, because <math>Z</math> lies on radical axis <math>TP</math> of the two circles, we have <cmath>ZM * ZA = ZN' * ZD.</cmath> Hence, <math>ZN'' = ZN'</math>, a contradiction since <math>D</math> and <math>D'</math> are distinct. We therefore conclude that <math>Z, N, D</math> are collinear, which gives the concurrency of <math>AM, XY</math>, and <math>DN</math>. This completes the problem. |
== Solution 3== | == Solution 3== | ||
− | Let <math>AM</math> and <math> | + | Let <math>AM</math> and <math>XY</math> intersect at <math>Z</math>. Because <math>\angle AMC = \angle BND = \angle APT = 90^\circ</math>, we have quadrilaterals <math>AMPT</math> and <math>DNPT</math> cyclic. Therefore, <math>Z</math> lies on the radical axis of the two circumcircles of these quadrilaterals. But <math>Z</math> also lies on radical axis <math>XY</math> of the original two circles, so the power of <math>Z</math> with respect to each of the four circles is all equal to <math>ZM * ZA</math>. Hence, <math>Z</math> lies on the radical axis <math>DN</math> of the two circles passing through <math>D</math> and <math>N</math>, as desired. |
+ | |||
+ | what is <math>T</math> here? | ||
+ | |||
+ | ==Discussion== | ||
+ | ''Lemma:'' The radical axis of two pairs of circles <math>O_1</math>, <math>O_2</math> and <math>O_3</math>, <math>O_4</math> are the same line <math>l</math>. Furthermore, <math>O_1</math> and <math>O_2</math> intersect at <math>A</math> and <math>B</math>, and <math>O_3</math> and <math>O_4</math> intersect at <math>C</math> and <math>D</math>. Then <math>A, B, C,</math> and <math>D</math> are concyclic. | ||
+ | |||
+ | The proof of this lemma is trivial using the argument in Solution 3 and applying the converse of Power of a Point. | ||
+ | |||
+ | Note that this Problem 1 is a corollary of this lemma. This lemma is an effective way to relate four circles, just as the radical center can relate three circles. | ||
+ | |||
+ | Solution 1 also gives a trivial lemma that can also be useful: | ||
+ | |||
+ | ''Lemma 2:'' Chords <math>AB</math> of <math>\omega_1</math> and <math>CD</math> of <math>\omega_2</math> intersect on the segment <math>XY</math> formed from the intersections of the two circles. Then <math>A, B, C, D</math> are concyclic. | ||
+ | |||
+ | Two ways to solve a problem, two different insights into circle geometry. That is cool, but more RADICAL! | ||
==See also== | ==See also== | ||
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] | ||
+ | {{IMO box|year=1995|before=First Question|num-a=2}} |
Latest revision as of 21:22, 4 July 2024
Problem
Let be four distinct points on a line, in that order. The circles with diameters
and
intersect at
and
. The line
meets
at
. Let
be a point on the line
other than
. The line
intersects the circle with diameter
at
and
, and the line
intersects the circle with diameter
at
and
. Prove that the lines
are concurrent.
Hint
Think about Radical Axis, Power of a Point and Radical Center.
Solution
Since is on the circle with diameter
, we have
and so
. We similarly find that
. Also, notice that the line
is the radical axis of the two circles with diameters
and
. Thus, since
is on
, we have
and so by the converse of Power of a Point, the quadrilateral
is cyclic. Thus,
. Thus,
and so quadrilateral
is cyclic. Let the name of the circle
be
. Then, the radical axis of
and the circle with diameter
is line
. Also, the radical axis of
and the circle with diameter
is line
. Since the pairwise radical axes of 3 circles are concurrent, we have
are concurrent as desired.
Solution 2
Let and
(a subsegment of
) intersect at
. Now, assume that
are not collinear. In that case, let
intersect the circle with diameter
at
and the circle through
at
.
We know that via standard formulae, so quadrilaterals
and
are cyclic. Thus,
and
are distinct, as none of them is
. Hence, by Power of a Point,
However, because
lies on radical axis
of the two circles, we have
Hence,
, a contradiction since
and
are distinct. We therefore conclude that
are collinear, which gives the concurrency of
, and
. This completes the problem.
Solution 3
Let and
intersect at
. Because
, we have quadrilaterals
and
cyclic. Therefore,
lies on the radical axis of the two circumcircles of these quadrilaterals. But
also lies on radical axis
of the original two circles, so the power of
with respect to each of the four circles is all equal to
. Hence,
lies on the radical axis
of the two circles passing through
and
, as desired.
what is here?
Discussion
Lemma: The radical axis of two pairs of circles ,
and
,
are the same line
. Furthermore,
and
intersect at
and
, and
and
intersect at
and
. Then
and
are concyclic.
The proof of this lemma is trivial using the argument in Solution 3 and applying the converse of Power of a Point.
Note that this Problem 1 is a corollary of this lemma. This lemma is an effective way to relate four circles, just as the radical center can relate three circles.
Solution 1 also gives a trivial lemma that can also be useful:
Lemma 2: Chords of
and
of
intersect on the segment
formed from the intersections of the two circles. Then
are concyclic.
Two ways to solve a problem, two different insights into circle geometry. That is cool, but more RADICAL!
See also
1995 IMO (Problems) • Resources | ||
Preceded by First Question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |